Find the limit of a function $$\lim_{x\to 3}\frac{\sqrt{3x} - 3}{\sqrt{2x-4} - \sqrt{2}}.$$
Letting 
$$F(x) = \frac{\sqrt{3x} - 3}{\sqrt{2x-4}-\sqrt{2}},$$
we have
$$F(x) = \frac{\sqrt{3}(\sqrt{x} - \sqrt{3})}{\sqrt{2}(\sqrt{x-2}-1)}.$$
Multiplying numerator and denominator by $\sqrt{x-2} + 1$,
$$F(x) = \frac{
(3)^{1/2} ((x(x-2)^{1/2})+(x)^{1/2}-(3(x-2)^{1/2})-(3)^{1/2})} 
{\sqrt{2}(x-3)}.$$
Dividing numerator and denominator by $x$ and substituting $3$ for $x$, I get $\frac{0}{\sqrt{2}} = 0$. 
Is it correct? My textbook does not have answer, one of the site gives the answer as $\frac{1}{\sqrt{2}}.$.
 A: Hint: $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$.
Applying the hint twice,
$$
F(x)=\sqrt{\frac32}\frac{\sqrt{x}-\sqrt3}{\sqrt{x-2}-1}=\sqrt{\frac32}\frac{x-3}{\sqrt{x}+\sqrt3}\,\frac{\sqrt{x-2}+1}{(x-2)-1}=\sqrt{\frac32}\frac{\sqrt{x-2}+1}{\sqrt{x}+\sqrt3},
$$
hence
$$
\lim\limits_{x\to3}\,F(x)=\sqrt{\frac32}\frac2{2\sqrt3}=\frac1{\sqrt{2}}.
$$
A: Your error is that after rationalizing the denominator, you cannot just plug in $3$ for $x$, because the denominator still evaluates to $0$ (not $\sqrt{2}$, as you seem to think). 
$$F(x) = \frac{\sqrt{3}}{\sqrt{2}}\cdot\left(\frac{\quad\frac{\sqrt{x}-\sqrt{3}}{x-3}\quad}{\quad\frac{\sqrt{x-2} - 1}{x-3}\quad}\right).$$
So
$$\lim_{x\to 3}F(x) = \frac{\sqrt{3}}{\sqrt{2}}\frac{\lim\limits_{x\to 3}\frac{\sqrt{x}-\sqrt{3}}{x-3}}{\lim\limits_{x\to 3}\frac{\sqrt{x-2}-1}{x-3}} = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{g'(3)}{h'(3)},$$
where $g(x) = \sqrt{x}$ and $h(x) = \sqrt{x-2}$.
Since $g'(x) = \frac{1}{2\sqrt{x}}$, $g'(3) = \frac{1}{2\sqrt{3}}$. Since $h'(x)=\frac{1}{2\sqrt{x-2}}$, then $h'(3) = \frac{1}{2}$.  So
$$\lim_{x\to 3}F(x) = \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{\frac{1}{2\sqrt{3}}}{\frac{1}{2}} = \frac{\quad\frac{1}{2}\quad}{\frac{\sqrt{2}}{2}}=\frac{1}{\sqrt{2}}.$$
